Waves_Week4and physics of wave_theory.pptx

its physics of waves

1. WAVES (One Dimensional) Wave Motion, Wave Equation and Solution, and Wave Velocity http://www.kettering.edu/~drussell/Demos/waves/wavemotion.html

2. passes by. http://www.kettering.edu/~drussell/Demos/waves-intro/waves-intro.html In longitudinal waves, the particles in a medium oscillate back and forth about their equilibrium positions but it is the disturbance which travels, not the individual particles in the medium. Longitudinal wave Transverse wave In a transverse wave the particle displacement is perpendicular to the direction of wave propagation. The particles do not move along with the wave; they simply oscillate up and down about their individual equilibrium positions as the wave http://www.kettering.edu/~drussell/Demos/waves/wavemotion.html Put simply, “a wave is a traveling disturbance”.

3. From Simple Harmonic Oscillation to Wave motion Lets try to find a link from what we have learnt till now so as to understand the phenomena of WAVES. Coupled oscillations form the natural link between simple harmonic motion (of a single particle) and wave motion ( of a continuous infinity of particles as in a medium )

4.  Oscillation of a single particle: One frequency ω  Oscillation of two coupled particles: Two normal frequencies ω , ω 1  Oscillation of n coupled particles: n normal frequencies ω , ω 1 …ω n- 1 y x  Oscillation of infinite no. of coupled particles (lattice/medium): T 𝑥(𝑡) = 𝐴𝑒 𝑖(𝜔 𝑜 𝑡+𝜑) From Simple Harmonic Oscillation to Wave motion

5. From Simple Harmonic Oscillation to Wave motion Consider a Two- Mass, Three- Spring System Longitudinal Motion (along x -axis): System equations Normal frequencies 𝑚 2 𝑑 𝑥 1 𝑑𝑡 2 + 𝑘𝑥 1 +𝑘(𝑥 1 −𝑥 2 ) = 𝑑𝑡 2 2 𝑚 𝑑 𝑥 2 + 𝑘𝑥 2 +𝑘(𝑥 2 −𝑥 1 ) = 𝜔 = 𝜔 , 3𝜔 𝑜 𝑜 𝑜 𝑤ℎ𝑒𝑟𝑒, 𝜔 = 𝑘 𝑚 Extrapolate to Multiple mass system Each additional mass adds another natural mode of vibration per axis of motion.

6. Lets us extend the concept of oscillations of a coupled spring mass system to a linear chain of springs and masses. Also, let there be long linear chain of identical springs of stiffness k connecting the identical blocks of mass m . Suppose we consider the limit when the number of springs and masses tends continuously to infinity. This kind of limiting case can be envisaged in an elastic rod (which is a continuous distribution of elasticity and inertia). If longitudinal disturbances are created on such a rod, they propogate much as in a linear chain of springs and masses, with the discrete system replaced by a continuous system. From Simple Harmonic Oscillation to Wave motion

7. Rod made of elastic substance A L Elastic Wave ©SB/SPK Disturbance in the rod

8. What happens to an elastic solid when it is compressed or stretched? Elasticity : Spring constant Stress is the internal force per unit area, associated with a strain. Strain is the relative change in shape or size of an object due to externally applied forces. (magnitude) 𝑆𝑡𝑟𝑒𝑠𝑠 = 𝐹 𝐴 𝜉 𝑆𝑡𝑟𝑎𝑖𝑛 = 𝐿 𝑆𝑡𝑟𝑒𝑠𝑠 𝑌 = 𝑆𝑡𝑟𝑎𝑖𝑛 (𝑌𝑜𝑢𝑛𝑔 ′ 𝑠 𝑀𝑜𝑑𝑢𝑙𝑢𝑠) 𝐹 = 𝑌𝐴 𝐿 𝜉 𝐹 = 𝑘𝜉 → 𝑆𝑝𝑟𝑖𝑛𝑔

9. ©SB/SPK i i- 1 i+1 Elasticity : Spring constant

10. i i- 1 i+1 Displacement 𝜉 𝑖 of i th mass Recall 𝑑 2 𝜉 𝑖 𝑚 = −𝑘(𝜉 𝑖 − 𝜉 𝑖+1 ) − 𝑘(𝜉 𝑖 − 𝜉 𝑖−1 ) 𝑑𝑡 2 dt 2 m i  k(  i  1   i )  k(  i   i  1 ) d 2  Displacement 𝜉 𝑖 of i th mass satisfies differential equation

11. 𝜕𝑡 𝜕𝜉(𝑥,𝑡) : variation of 𝜉(𝑥, 𝑡) with t while x is kept constant 𝜕𝜉(𝑥,𝑡) : variation of 𝜉(𝑥, 𝑡) with x while t is kept constant 𝜕𝑥 Let, a : separation between the masses a  𝛿𝑥 where 𝛿𝑥 → In the Continuum limit 𝜉 𝑖 → 𝜉(𝑥, 𝑡) 𝜉 𝑖+1 → 𝜉(𝑥 + 𝛿𝑥, 𝑡) 𝜉 𝑖−1 → 𝜉(𝑥 − 𝛿𝑥, 𝑡) i i- 1 i+1 𝜉 is a function of two continuous variable x and t : 𝜉 = 𝜉(𝑥, 𝑡) Notation of partial derivatives

12. Taylor series expansion A Taylor series is a representation of a function as an infinite sum of terms that are calculated from the values of the function's derivatives at a single point.  i  1 ( x , t )   i ( x , t )   ( x   x , t )   ( x , t ) (  x ) 2    1  2   x 2  x 2   x and ( x , t )   ( x , t )   ( x   x , t )   (   x ) i (  x ) 2 i  1  ( x , t )    2     1  x 2  x 2 dt 2   i  1 ) m i  k (  i  1   i )  k (  i d 2  𝜉 𝑖 → 𝜉(𝑥, 𝑡) 𝜉 𝑖+1 → 𝜉(𝑥 + 𝛿𝑥, 𝑡) 𝜉 𝑖−1 → 𝜉(𝑥 − 𝛿𝑥, 𝑡) Subtract: k [(1) – (2)] → (2) → (1) 2 m  k (  x )  t 2  x 2  2   2  2 m  x 2 ( x )  t 2  2   k  2  

13. Longitudinal wave in elastic rod  x YA k  m   A  x c s : wave velocity  2   Y  2   t 2   x 2 s c 2  2   1  2   x 2  t 2 Y : Young’s modulus A : Cross sectional area   Mass density  t 2 m  x 2 Wave equation: 2  2   k  2  (  x ) We have: 𝑐 𝑠 = 𝑌 𝜚

14. Transverse vibrations in strings

15. Loaded string case  2  1 m m m r y - y r- 1 y - y r r+1 y r y r- 1 y r+1 a a

16. a a n=1 A 1 A = A 2 = a a n=2 a A 1 A 2 A = A 3 = m m Anti-phase / Breathing mode Loaded string case a In- phase / Pendulum mode n=2 a m A 1 A 2 A = 3 A = m a m

17. Loaded string case a light string supporting n equal masses m spaced at equal distance a along its length The string is fixed at both ends; it has a length ( n +1) a and a constant tension T exists at all points and all times in the string. Small SHO of the masses are allowed in only one plane and the problem is to find the frequencies of the normal modes and the displacement of each mass in a particular normal mode.

18. y r - y r+1 The equation of motion of this mass may be written by considering the components of the tension directed towards the equilibrium position. The r th mass is pulled downwards towards the equilibrium position by a force T sin  1 , due to the tension on its left and a force T sin  2 due to the tension on its right. Loaded string case Equation of motion:  y r  1  2 y r  y r  1  dt 2 d 2 y dt 2 d 2 y m r   T (sin  1  sin  2 ) so ,  r   y  r        T ma T a  y r  1  a r  y r  1 y  y r T cos  1  2 m m m y r - y r- 1 y r- 1 y r+1 a a T T cos  2 T sin  2 y r T sin  1 T  1

19. We know, In a given mode all masses oscillate with the same mode frequency  , so all y r ’s have the same time dependence. However, the transverse displacement y r also depends upon the value of r i.e., the position of the r th mass on the string. This means that y r is a function of two independent variables , the time t and the location of r on the string. If we use the separation a ≈  x and let  x  0, the masses become closer and we can consider positions along the string in terms of a continuous variable x and any transverse displacement as y(x,t) , a function of both x and t . In this case, partial derivative notations can be used. 20 Loaded string case

20. If we now locate the transverse displacement y r at a position x = x r along the string then, we get: r   t 2  t 2  2 y  2 y where y is evaluated at x = x r , and now, as a =  x  , ( in the Continuum limit) we may write x r = x, x r+1 = x +  x and x r- 1 = x -  x with y r (t)  y(x,t), y r+1 (t)  y(x +  x, t) and y r- 1 (t)  y(x -  x, t) Taylor series expansion Loaded string case

21. = 𝑇 𝑚 𝛿𝑥 𝜕𝑦 1 𝛿𝑥 𝜕𝑥 + 2 𝜕𝑥 2 𝛿𝑥 − 𝛿𝑥 𝜕𝑦 1 𝛿𝑥 𝜕𝑥 − 2 2 𝜕 2 𝑦 𝜕𝑥 2 𝛿𝑥 So, the eq. of motion becomes after the substitution: = 𝜕𝑡 2 𝑚 𝑟+1 𝜕 2 𝑦 𝑇 𝑦 − 𝑦 𝑟 𝑎 − 𝑦 − 𝑦 𝑟 𝑟−1 𝑎 2 𝜕 2 𝑦    a m r   T  r r  1  r r  1  dt 2 a d 2 y  y  y y  y y r (t)  y(x,t), y r+1 (t)  y(x +  x, t) y r- 1 (t)  y(x -  x, t) Loaded string case 𝑠𝑜, 𝜕 2 𝑦 𝑇 𝜕𝑡 2 = 𝑚 2 𝜕 2 𝑦 𝛿𝑥 𝑇 𝜕 2 𝑦 𝛿𝑥 𝜕𝑥 2 = 𝑚 𝛿𝑥 𝜕𝑥 2  x 2   1 2  2    x   x , t    ( x )   x  x  2 (   x )  ......

22. If we now write m =   x where,  is the linear density (mass per unit length) of the string, the masses must  as  x  to avoid infinite mass density. Thus, we have: 𝜕 2 𝑦 𝑇 𝜕 2 𝑦 𝜕𝑡 2 = 𝜌 𝜕𝑥 2 This is the WAVE EQUATION . T/  has the dimension of the square of the velocity, the velocity with which the waves; i.e., the phase of oscillation, is propagated. The solution for y at any particular point along the string is always that of a harmonic oscillator. Q: How do we know that c s is the wave velocity? Ans: As of now we cannot say if c s is wave velocity of particle velocity or some other velocity. Wait for it… 𝑠 1 𝜕 2 𝑦 𝜕 2 𝑦 𝑐 2 𝜕𝑡 2 = 𝜕𝑥 2 Loaded string case 𝜕 2 𝑦 𝑇 𝜕𝑡 2 = 𝑚 2 𝜕 2 𝑦 𝛿𝑥 𝑇 𝜕 2 𝑦 𝛿𝑥 𝜕𝑥 2 = 𝑚 𝛿𝑥 𝜕𝑥 2

23. Wave equation and solution

24.  c  t 2  x 2  2  2  2  Wave equation General Solution Solutions Wave equation and Solution

25. Wave equation and Solution 𝐿𝑒𝑡 𝑢 = 𝑐𝑡 − 𝑥 𝑣 = 𝑐𝑡 + 𝑥 ‘Chain rule’

26. Physical significance of  = f(ct- x) Disturbance: ξ = f(x,t) If we photograph the wave at t=0 : After a time t , the pulse has moved a distance ct Introduce frame S’ which travels with the pulse S x 1 S’ x 1 ’ ct S ξ = f(x’) x 1 ’ x’ x ξ = f(x,t) x 2 x 1 ’ = x 2 – ct  In general, x’ = x – ct For someone at rest in S, the pulse moving to right can be denoted by: ξ = f(x’) = f(x- ct) Verify that the behavior of ξ = f(x- ct) and ξ = f(ct- x) are similar.

27. x  wave moving to the left Physical significance of  = f(ct+x)

28. If ξ is displacement of a SHO at position x and time t , we can express it as: The bracket in  = f(ct- x) has dimension of length.   a cos(  t   ) For this function to be a sine or cosine , its argument must have dimensions of radians. 𝜆 2𝜋 𝜉 = 𝑎 cos( 𝜔𝑡 − 𝜑) = 𝑎 cos (𝑐𝑡 − 𝑥) So, we can write the solution as : where, 2𝜋 2𝜋 𝑐 = 𝜔 = 2𝜋ν; 𝑥 = 𝜑 𝜆 𝜆  : frequency of oscillation More on solution to Wave equation Recollect the question: How do we know that c s (or c ) is the wave velocity? 𝜔 𝑐 = 𝑘 Here, , which is nothing but the wave velocity! 𝜉 = 𝑎 cos 2𝜋 𝜆 (𝑐𝑡 − 𝑥)  = f(ct- x) , which satisfies the wave equation, can be written in the form:

29. x is a point along the x - axis. y and z are not part of the equation because the wave's magnitude and phase are the same at every point on any given y- z plane. This equation defines what that magnitude and phase are. 𝜉 = 𝑎 cos 2𝜋 𝜆 (𝑐𝑡 − 𝑥) Q: Here, what is a plane composed of? https://commons.wikimedia.org/wiki/File:Plane_Wave_3D_Animation_300x216_255Colors.gif Plane wave

30. For 𝑥 = 𝑛𝜆 𝜆: Wavelength the pattern repeats 𝑘 = 2𝜋 𝜆 : Wavenumber 𝜆 2𝜋𝑐 = 𝜔 = 2𝜋𝜈 2  c   ; 2  x     The number of wavelengths that exist over a specified distance c : wave velocity 𝑐 = 𝜈𝜆  1 c      : period of oscillation Other important relations 𝜔 = 𝑘𝑐 More on solution to Wave equation For monochromatic wave in a non- dispersive medium    k Slope ( c )  phase velocity of the wave What is dispersion? Wait for it… Plot of dispersion relation

31. Standing Waves

32. Boundary conditions Standing waves on a Stretched string l 𝑥 = 0, 𝜓 = ⇒ 𝐴 + 𝐵 = 𝜓 = −𝐴2𝑖 exp 𝑖𝜔𝑡 sin 𝑘𝑥

33. ( n- 1 ) nodes between boundaries Standing waves on a Stretched string 𝜓 = −𝐴2𝑖 exp 𝑖𝜔𝑡 sin 𝑘𝑥 𝜓 = −𝐴2𝑖 exp 𝑖𝜔𝑡 sin 𝑛𝜋𝑥 𝑙 (kl = nπ)

34. This is standing wave Wave in Time and Space Frames These are Progressive Wave 𝜓 = −𝐴2𝑖 exp 𝑖𝜔𝑡 sin 𝑘𝑥 T t Visualize a Progressive wave at a given point in SPACE

35. x Wave in Time and Space Frames Visualize a Progressive wave at a given point in TIME 𝑐 = 𝜆 = 2𝜋𝜆 = 𝜔 𝑇 2𝜋𝑇 𝑘